Irodov 3.38: Half Again Higher at the Center – Amit & Deepali Sinha Foundation

Problem Irodov 3.38

A ball of radius $R$ carries a charge $q$ spread evenly through its volume. Take the permittivity to be 1 everywhere, inside the ball and outside it. Find the electric potential

at the center of the ball;
inside the ball, as a function of the distance $r$ from its center.

The ball of radius R with charge q spread evenly through it (drawn for q > 0), its center O, and a point P inside at distance r from the center. — The ball of radius $R$ with charge $q$ spread evenly through it (drawn for $q > 0$ ), its center $O$ , and a point $P$ inside at distance $r$ from the center.

Given

$q$ uniform through the ball
radius $R$
$\varepsilon = 1$ throughout

Find

(a)

\varphi

at the center · (b)

\varphi(r)

for

r \le R

Solution

Step 1The field inside and outside

The charge has spherical symmetry, so the field is radial and has the same strength all over any sphere centered at $O$ . Gauss’s law on a sphere of radius $r$ then needs only the charge that sphere encloses. Inside the ball that is the fraction $r^3/R^3$ of $q$ :

\begin{align*} E\cdot 4\pi r^2 &= \frac{q}{\varepsilon_0}\,\frac{r^3}{R^3} \\ E &= \frac{q\,r}{4\pi\varepsilon_0 R^3} \qquad (r \le R) \tag{1}\end{align*}

Outside, the sphere encloses all of $q$ :

\begin{align*} E\cdot 4\pi r^2 &= \frac{q}{\varepsilon_0} \\ E &= \frac{q}{4\pi\varepsilon_0 r^2} \qquad (r \ge R) \tag{2}\end{align*}

Step 2The potential at the surface

Take the potential to be zero far away. The potential at distance $r$ is then the work per unit charge done by the field on the way out to infinity, $\varphi(r) = \int_r^\infty E\,dr'$ . At the surface only the outside field (2) enters:

\varphi(R) = \int_R^\infty \frac{q\,dr'}{4\pi\varepsilon_0 r'^2} = \frac{q}{4\pi\varepsilon_0}\left[-\frac{1}{r'}\right]_R^\infty = \frac{q}{4\pi\varepsilon_0 R} \tag{3}

Step 3Integrating inward to the center

From a point inside, the path out first crosses the ball, where (1) holds, and then continues from $R$ outward, which adds (3):

\begin{aligned} \varphi(r) &= \int_r^R \frac{q\,r'\,dr'}{4\pi\varepsilon_0 R^3} + \varphi(R)\\ &= \frac{q}{4\pi\varepsilon_0 R^3}\cdot\frac{R^2 - r^2}{2} + \frac{q}{4\pi\varepsilon_0 R}\\ &= \frac{q}{8\pi\varepsilon_0 R}\left(3 - \frac{r^2}{R^2}\right) \end{aligned}

At the center, $r = 0$ , the bracket is 3:

\varphi_0 = \frac{3q}{8\pi\varepsilon_0 R}

(4)

Taking $\varphi_0$ out of the bracket, the potential at any depth inside is

\varphi(r) = \varphi_0\left(1 - \frac{r^2}{3R^2}\right)\qquad (r \le R)

(5)

The field E(r) (crimson, top) and the potential φ(r) (teal, bottom) of the ball. Inside, E grows linearly and φ is a downward parabola, flat at the center; outside, both are those of a point charge. — The field $E(r)$ (crimson, top) and the potential $\varphi(r)$ (teal, bottom) of the ball. Inside, $E$ grows linearly and $\varphi$ is a downward parabola, flat at the center; outside, both are those of a point charge.

Check

The center value follows independently, without the field, by adding the potentials of thin shells: a shell of radius $r$ and thickness $dr$ holds the charge $3q r^2\,dr/R^3$ and gives at the center the potential of that charge at distance $r$ ,

\varphi_0 = \int_0^R \frac{3q r^2\,dr}{4\pi\varepsilon_0 R^3\,r} = \frac{3q}{4\pi\varepsilon_0 R^3}\cdot\frac{R^2}{2} = \frac{3q}{8\pi\varepsilon_0 R}

which is (4).

Remark

The center sits at $3/2$ of the surface potential. The same profile fixes the ball’s self energy: half the integral of $\rho\varphi$ over the volume, with this $\varphi(r)$ , gives $W = 3q^2/(20\pi\varepsilon_0 R)$ . And because $\varphi$ inside is a parabola, a small opposite charge free to move through the ball (Thomson’s old picture of an atom) feels a force proportional to its distance from the center and oscillates harmonically about it.